Does my definition of an exponential-like function make any sense? The definition of the exponential function is based on an infinite series
$$
e^x = \sum_{k=0}^{\infty}\frac{x^k}{k!}
$$
To make things more complicated, we could replace the factorial with the Gamma function
$$
e^x = \sum_{k=0}^{\infty}\frac{x^k}{\Gamma(k+1)}
$$
The advantage of the Gamma function is that it is defined also for continuous values, so we are tempted to replace the infinite sum with an integral
$$
\epsilon^x = \int_{k=0}^{\infty}\frac{x^k}{\Gamma(k+1)}dk
$$
It turns out that this definition is somewhat similar to the exponential function, but not quite the same. The constant $\epsilon$ is computed by Wolfram Alpha as
$$
\epsilon = \int_{k=0}^{\infty}\frac{1^k}{\Gamma(k+1)}dk = 2.2665345...
$$
but from the integral we have
$$
\epsilon^0 = \int_{k=0}^{\infty}\frac{0^k}{\Gamma(k+1)}dk = 0
$$ 
which seems to be in contradiction to the rule $a^0 = 1$ for every $a\neq 0$. 
So is my integral definition for $\epsilon^x$ consistent at all? How does the integral behave for $x \rightarrow \pm\infty$ ? Is there any literature about this kind of integral and it's relation to the definition of the exponential function?
 A: Your integral $\ \int_{k=0}^{\infty}\frac{x^k}{\Gamma(k+1)}dk\ $ is perfectly well-defined for $\ x \ge 0\ $.  For $\ x < 0\ $ you run into the problem that there is no unique definition for $\ x^k\ $ when $\ k\ $ is not an integer. In general, there are several possible values, $\ \vert x\vert^k\,e^{(2n\pi+1)k i}\ $, to choose for it, and these becone infinite in number if $\ k\ $ is irrational.  Here, $\ n\ $ can be any integer, but if you simply choose one particular value for it ($\ n=0\ $, for instance), then the integral will be well-defined for $\ x < 0\ $ as well, although it will typically be complex-valued in that case.
You have no grounds, however, for assuming that the integral will have the  value  $\ \ \left(\int_{k=0}^{\infty}\Gamma(k+1)^{-1}dk\right)^x\ $, as a couple of commenters have already pointed out, and, in fact, it doesn't, except when $\ x=1\ $, so the apparent inconsistency which puzzles you has resulted from this assumption, rather than from any problem with the definition of the integral itself.
You can show that the integral is well-defined for
$\ x \ge 0\ $, and obtain bounds for it, by observing that $ \frac{x^k}{\Gamma(k+1)}\ $ is bounded and continuous for $ k\ge0\ $, and for $\ 0\le i\le k\le i+1 $ we have $\ i\,! \le \Gamma\left(k+1\right) \le \left(i+1\right)\,!\ $ and $\ x^i \le x^k \le x^{i+1}\ $ if $\ x \ge 1\ $, or $\ x^{i+1} \le x^k \le x^i\ $ if $ x < 1\ $. Therefore $ \frac{x^i}{\left(i+1\right)\,!} \le \frac{x^k}{\Gamma\left(k+1\right)} \le \frac{x^{i+1}}{i\,!}\ $ if $\ x \ge 1\ $, and $ \frac{x^{i+1}}{\left(i+1\right)\,!} \le \frac{x^k}{\Gamma\left(k+1\right)} \le \frac{x^i}{i\,!}\ $ if $ x < 1 $.
Thus, for $\ x\ge 1\ $,
$$\sum_{i=0}^r \frac{x^i}{\left(i+1\right)\,!}\le \ \int_{k=0}^{r+1}\frac{x^k}{\Gamma(k+1)}dk\ \le \sum_{i=0}^r \frac{x^{i+1}}{i\,!}\le x\,e^x\ \ ,$$
so the integral converges as $\ r\rightarrow\infty\ $, is bounded below by $\ \frac{e^x-1}{x}\ $ and above by $\ x\,e^x\ $.
On the other hand, for $\ x < 1\ $,
$$\sum_{i=0}^r \frac{x^{i+1}}{\left(i+1\right)\,!}\le \ \int_{k=0}^{r+1}\frac{x^k}{\Gamma(k+1)}dk\ \le \sum_{i=0}^r \frac{x^i}{i\,!}\le e^x\ \ ,$$
so the integral again converges as $\ r\rightarrow\infty\ $, is bounded below by $\ e^x-1\ $ and above by $\ e^x\ $.
